4 ms·

Most intro courses on QM that I've seen cover exclusively problems of operators with continuous spectra (like position and momentum). For those, the mathematics

Most intro courses on QM that I've seen cover exclusively problems of operators with continuous spectra (like position and momentum). For those, the mathematics of L2 space is critical (and, in my experience, woefully neglected). Because multiple observables are rarely treated, tensor products and entanglement are less important. For intro to QIT, thoroughly understanding the C^2 space and tensor products is critical. This book appears to be useful for that. Unfortunately, it can be hard to make the jump to the former way of looking at things, without someone explicitly pointing out how to think of L2 as a Hilbert space. Maybe this is just intuitive to some, but it wasn't for me. And the converse seems to be true, too: you can apparently master the former without having a clue about the latter. As Scott Aaronson points out: > Today, in the quantum information age, the fact that all the physicists had to learn quantum this way seems increasingly humorous. For example, I've had experts in quantum field theory -- people who've spent years calculating path integrals of mind-boggling complexity -- ask me to explain the Bell inequality to them. That's like Andrew Wiles asking me to explain the Pythagorean Theorem. https://www.scottaaronson.com/democritus/lec9.html https://www.scottaaronson.com/democritus/lec9.html

kgwgk 8y ago
This reminds me of Peter Woit, a theoretical physicist who has written a book on "Quantum Theory, Groups and Representations", asking if QM is a probabilistic theory. https://www.math.columbia.edu/~woit/wordpress/?p=10533 https://www.math.columbia.edu/~woit/wordpress/?p=10533 By the way, this example illustrates the point I made in another comment: you can have a long career in physics and even become professor of theoretical physics without ever caring about the interpretation of QM. Edit: I can't resist copying the opening of a recent comment in that thread. ---- John Baez says: September 18, 2018 at 4:05 pm Peter Woit wrote: "The state of the world is described at a fixed time by a state vector, which evolves unitarily by the Schrodinger equation. No probability here." And perhaps no physics here, either, unless we say how the state of the world is described by that vector: that is, how we can use the vector to make predictions of experimental results. (A long and insightful comment by Baez follows)
jostylr 8y ago
To link the two approaches, it can be useful to understand the wave function as an object that takes values in a tensor space of C^n. In particular, if we had two particles, both spin-1/2 (C^2), then at the position configuration point {x,y}, the wave function takes a value in C^2 tensor C^2 where the tensoring is not done over the particle numbering 1 and 2 (which does not exist), but done over the set {x,y}. This is how one naturally leads into the choices of bosons vs. fermions. To say it another way, the configuration of the system remains rooted in position in physical space and it is the value space of the wave function where spin resides. This is made most clear in Bohmian mechanics, where the position of the particle is always defined in the theory, but whether a particle will end up in the spin-z up or down regions does depend on the experimental setup. One can arrange experiments where the particle, due to symmetry for example, will always go up given the same starting point even if one flips the magnetic field in the Stern-Gerlach device so that the experimental conclusion would be spin up in one scenario and spin down in the other. Same exact path happens in both cases, but the spin value conclusion is the opposite. That is to say, while position is determined ahead of time, spin is not. Spin is not a real property of the particle; it is a property of the wave function.

kgwgk 8y ago

This reminds me of Peter Woit, a theoretical physicist who has written a book on "Quantum Theory, Groups and Representations", asking if QM is a probabilistic theory. https://www.math.columbia.edu/~woit/wordpress/?p=10533 https://www.math.columbia.edu/~woit/wordpress/?p=10533 By the way, this example illustrates the point I made in another comment: you can have a long career in physics and even become professor of theoretical physics without ever caring about the interpretation of QM. Edit: I can't resist copying the opening of a recent comment in that thread. ---- John Baez says: September 18, 2018 at 4:05 pm Peter Woit wrote: "The state of the world is described at a fixed time by a state vector, which evolves unitarily by the Schrodinger equation. No probability here." And perhaps no physics here, either, unless we say how the state of the world is described by that vector: that is, how we can use the vector to make predictions of experimental results. (A long and insightful comment by Baez follows)

jostylr 8y ago

To link the two approaches, it can be useful to understand the wave function as an object that takes values in a tensor space of C^n. In particular, if we had two particles, both spin-1/2 (C^2), then at the position configuration point {x,y}, the wave function takes a value in C^2 tensor C^2 where the tensoring is not done over the particle numbering 1 and 2 (which does not exist), but done over the set {x,y}. This is how one naturally leads into the choices of bosons vs. fermions. To say it another way, the configuration of the system remains rooted in position in physical space and it is the value space of the wave function where spin resides. This is made most clear in Bohmian mechanics, where the position of the particle is always defined in the theory, but whether a particle will end up in the spin-z up or down regions does depend on the experimental setup. One can arrange experiments where the particle, due to symmetry for example, will always go up given the same starting point even if one flips the magnetic field in the Stern-Gerlach device so that the experimental conclusion would be spin up in one scenario and spin down in the other. Same exact path happens in both cases, but the spin value conclusion is the opposite. That is to say, while position is determined ahead of time, spin is not. Spin is not a real property of the particle; it is a property of the wave function.